Question

(1) The following diagram represents a titration curve of histidine as pH increases; pKa = 1.82 represents the terminal carboxylic acid group. pK, represents the terminal amino group: 6.0 represents the R-group, and pK, E 9.17 7.59 pH 182 3.0 20 10 H (equivalents) At what point on the diagram is histidine predominantly present as the following species? COO A, at pH 9.17 H2 H CH (2) What is the net charge on the following polypeptide at pH 11? Use the pK, table on the last Equations and Constants page of the exam. C. 0 D. +1 E. +2 Amino Acid pK Glycine 9.60 9.69 9.62 9.60 9.60 9.15 10.43 9.21 9.13 9.39 2.32 2.36 Leucine ucine 2.21 Threonine Methionine Phenylalanine 183 Asparagine Glutamine Proline Cysteine Histidine Aspartic acid Glutamic acid 9.13 10.60 10.78 9.17 9.82 9.67 9.11 8.95 9.04 2.17 1.99 1.71 1.82 6.00 2.19 10.07 10.79 2.18 2.17 Arginine

Answer 1

1. (C) between pH 6 and 9.7

Histidine has a basic side chain or R group which introduces an extra positive charge. So the neutral form exists under more basic conditions when the extra positive charge has been neutralized, hence it is found more in between the pH range 6 to 9.7.

2. (B) The amino acids that constitute the polypeptide includes tyrosine, threonine, glutamine, tryptophan and isoleucine.The only available alpha-amino and alpha-carboxylic acid group are on tyrosine, glutamine and isoleucine, the rest of the them were used for peptide bond formation.

If the pH is higher than the pKa value, it favors deprotonation and vice-versa.

The amino acid shows a deprotonizable amine group (as pKa 9.11 of amine group in tyrosine is less than pH 11) on the tyrosine substituent, with a neutral charge. It shows a deprotonizable amine group (as pKa 9.13 of amine group is less than pH 11) on the glutamine substituent with a neutral charge. It shows a deprotonizable carboxylic acid group (as pKa 2.36 of carboxylic acid group is less than pH 11) on the isoleucine substituent and hence has a negative charge. Hence overall the peptide has a negative charge of -1.